Question

a 10g bullet travelling at 20m/s strikes and remains embedded in a 2kg target with which is originally at rest but free to move. At what speed does the target move off?

Answer 1

heya....


Mass of bullet= 10 g
SPEED of bullet=200 m/s


Mass in which bullet embedded=2kg
let the Velocity of block be "v"


Applying conservation of mass, 
10(200)=(2000g) v

=> v=1 m/s



tysm...#gozmit

Answer 2

Hey mate!

Here's your answer!!

Mass of bullet, m = 10 g = 0.01 kg

Speed of bullet, v = 200 m/s

Mass of target, M = 2 kg

Combine speed after hitting target = V

Before collision, momentum of the system is
= mv = 2

After collision, the bullet and target move together.
Their combined mass is (m+M) = 2.01 kg

So, the momentum of the system after collision is = 2.01V

Applying conservation of momentum,

Momentum of the system before collision = momentum of the system after collision

=> 2 = 2.01 V

=> V = 0.99 m/s

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